14 13 One can show that this tensor is traceless. Indeed by using the eq. (??) and the fact that the spinor is symmetric we see that the tensor A is traceless. The inverse transformation is defined by ψ A1...A j B 1...B j = 1 j Ai 1...i j σ i1 A 1 B 1 σ i j A j B j (.74) The double covering homomorphism Λ : S U() S O(3) (.75) is defined as follows. Let U S U(). Then the matrices Uσ i U 1 satisfy all the properties of the Pauli matrices and are therefore in the algebra su(). Therefore, Uσ i U 1 = Λ j i(u)σ j. (.76) That is, Λ i j(u) = 1 tr σi Uσ j U 1 (.77) The matrix Λ(U) depends on the matrix U and is, in fact, in S O(3). For infinitesimal transformations U = exp(x k y k ) = I + i σ ky k + (.78) the matrix Λ is Λ i j = δ i j + ε i jk y k + = [ exp(x k y k ) ] i j (.79) It is easy to see that So, in short where Λ(I) = Λ( I) = I. (.80) Λ(exp(X i y i )) = exp(λ(x i )y i ) (.81) [Λ(σ k )] i j = iε i k j (.8) 3 Invariant Vector Fields Let y i be the canonical coordinates on the group S U() ranging over ( π, π). The position of the indices on the coordinates will be irrelevant, that is, y i = y i. We introduce the radial coordinate r = y = y i y i, so that the geodesic distance su.tex; November 9, 011; 10:07; p. 13

22 1 where T = T a T a. This means that exp[t 0 ] exp[t(ω)] = exp(tt ) exp[t(ω)] (3.164) The Killing vector fields K a ± are divergence free, which means that they are anti-self-adjoint with respect to the invariant measure dvol (ω). that is, Thus the Laplacian 0 is self-adjoint with respect to the same measure as well. Let be the total connection on a vector bundle V realizing the representation G. Let us define the one-forms A = σ a +G a = G a Y a µdx µ. (3.165) and F = da + A A. Then, by using the above properties of the right-invariant one-forms we compute where F = 1 F abσ a + σ b + = 1 F aby a µy b νdx µ dx ν (3.166) F ab = ε c abg c (3.167) Thus A is the Yang-Mills curvature on the vector bundle V with covariantly constant curvature. The covariant derivative of a section of the bundle V is then (recall that a = K + a ) a ϕ = (K + a + G a )ϕ (3.168) and the covariant derivative of a section of the endomorphism bundle End(V) along the right-invariant basis is then Therefore, we find a Q = K + a Q + [G a, Q] (3.169) a G b = ε c bag c + [G a, G b ] = 0. (3.170) Then the derivatives along the left-invariant vector fields are K a ϕ = (K a + B a )ϕ (3.171) where K a Q = K a Q + [B a, Q] (3.17) B a = D a b G b (3.173) su.tex; November 9, 011; 10:07; p. 1

23 The Laplacian takes the form = a a = (K + a + G a )(K + a + G a ) = 0 + G a K + a + G (3.174) where G = G i G i. We want to rewrite the Laplacian in terms of Casimir operators of some representations of the group S U(). The covariant derivatives a do not form a representation of the algebra S U(). The operators that do are the covariant Lie derivatives. The covariant Lie derivatives along a Killing vector ξ of sections of this vector bundle are defined by L ξ = ξ 1 σa +( b ξ)ε bc ag c (3.175) By denoting the Lie derivatives along the Killing vectors K ± a by K ± a this gives for the right-invariant and the left-invariant bases K + a = L K + a = a + G a = K + a + G a. (3.176) K a = L K a = K a B a = K a. (3.177) It is easy to see that these operators form the same algebra S U() S U() [K + a, K + b ] = εc abk + c (3.178) [K a, K b ] = εc abk c (3.179) [K + a, K b ] = 0 (3.180) Now, the Laplacian is given now by the sums of the Casimir operators = K G (3.181) where K = 1 K K (3.18) and K ± = K ± a K ± a. 4 Heat Kernel on S 3 Our goal is to evaluate the heat kernel diagonal. Since it is constant we can evaluate it at any point, say, at the origin. We use the geodesic coordinates y i defined su.tex; November 9, 011; 10:07; p.

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